How To Solve Systems With 3 Variables

Solving systems of equations with three variables might seem daunting at first, but with a systematic approach, it becomes a manageable task. These systems, often encountered in algebra and various real-world applications, involve finding the values of three unknowns that satisfy a set of three linear equations. This comprehensive guide will walk you through the methods, strategies, and underlying concepts to master this skill.

Understanding Systems of Equations with Three Variables

A system of equations with three variables, typically denoted as x, y, and z, consists of three linear equations. Each equation represents a plane in a three-dimensional space, and the solution to the system is the point where all three planes intersect. This intersection point, if it exists, provides the values for x, y, and z that satisfy all three equations simultaneously.

A general form of such a system looks like this:

• ax + by + cz = d
• ex + fy + gz = h
• ix + jy + kz = l

Where a, b, c, d, e, f, g, h, i, j, k, and l are constants, and x, y, and z are the variables we need to solve for.

Types of Solutions

When dealing with systems of equations, it's important to know the types of solutions you might encounter:

• Unique Solution: The system has exactly one solution, meaning the three planes intersect at a single point.
• No Solution: The system is inconsistent, meaning the planes do not intersect at any common point. This can occur when the planes are parallel or intersect in a way that no single point lies on all three.
• Infinite Solutions: The system is dependent, meaning the planes intersect along a line or are the same plane. In this case, there are infinitely many points that satisfy all three equations.

Methods to Solve Systems of Equations with Three Variables

Several methods can be used to solve systems of equations with three variables, including:

1. Substitution Method
2. Elimination Method
3. Gaussian Elimination
4. Matrix Methods (using inverse matrices or Cramer's Rule)

We will explore each of these methods in detail, providing examples and step-by-step instructions.

1. Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equations to reduce the number of variables.

Steps:

1. Solve one equation for one variable: Choose the easiest equation and solve for one of the variables.
2. Substitute: Substitute the expression obtained in step 1 into the other two equations. This will result in a system of two equations with two variables.
3. Solve the new system: Solve the two-variable system using either substitution or elimination.
4. Back-substitute: Once you have the values of two variables, substitute them back into one of the original equations to find the value of the third variable.

Example:

Solve the following system:

• x + y + z = 6 ...(1)
• 2x - y + z = 3 ...(2)
• x + 2y - z = 2 ...(3)

Solution:

1. Solve equation (1) for x:

   x = 6 - y - z

2. Substitute this expression for x into equations (2) and (3):

   • 2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9 ...(4)
   • (6 - y - z) + 2y - z = 2 => 6 + y - 2z = 2 => y - 2z = -4 ...(5)

3. Solve the new system (4) and (5):

   Solve equation (5) for y:

   y = 2z - 4

   Substitute this into equation (4):

   -3(2z - 4) - z = -9 => -6z + 12 - z = -9 => -7z = -21 => z = 3

   Now, substitute z = 3 back into y = 2z - 4:

   y = 2(3) - 4 => y = 6 - 4 => y = 2

4. Back-substitute y = 2 and z = 3 into x = 6 - y - z:

   x = 6 - 2 - 3 => x = 1

Thus, the solution is x = 1, y = 2, and z = 3.

2. Elimination Method

The elimination method involves adding or subtracting multiples of equations to eliminate one variable at a time, simplifying the system.

Steps:

1. Choose a variable to eliminate: Look for equations where the coefficients of one variable are the same or easily made the same.
2. Eliminate the variable: Multiply one or both equations by a constant so that the coefficients of the chosen variable are additive inverses. Then, add the equations to eliminate the variable.
3. Repeat: Repeat steps 1 and 2 to eliminate the same variable from a different pair of equations. This will give you a system of two equations with two variables.
4. Solve the new system: Solve the two-variable system using either substitution or elimination.
5. Back-substitute: Once you have the values of two variables, substitute them back into one of the original equations to find the value of the third variable.

Example:

Solve the following system:

• x + y + z = 6 ...(1)
• 2x - y + z = 3 ...(2)
• x + 2y - z = 2 ...(3)

Solution:

1. Eliminate y from equations (1) and (2):

   Add equation (1) and equation (2):

   (x + y + z) + (2x - y + z) = 6 + 3 => 3x + 2z = 9 ...(4)

2. Eliminate y from equations (1) and (3):

   Multiply equation (1) by -2:

   -2(x + y + z) = -2(6) => -2x - 2y - 2z = -12

   Add this modified equation to equation (3):

   (-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 => -x - 3z = -10 => x + 3z = 10 ...(5)

3. Solve the new system (4) and (5):

   Multiply equation (5) by -3:

   -3(x + 3z) = -3(10) => -3x - 9z = -30

   Add this modified equation to equation (4):

   (3x + 2z) + (-3x - 9z) = 9 + (-30) => -7z = -21 => z = 3

   Substitute z = 3 back into equation (5):

   x + 3(3) = 10 => x + 9 = 10 => x = 1

4. Back-substitute x = 1 and z = 3 into equation (1):

   1 + y + 3 = 6 => y + 4 = 6 => y = 2

Thus, the solution is x = 1, y = 2, and z = 3.

3. Gaussian Elimination

Gaussian elimination is a systematic method used to solve systems of linear equations by transforming the augmented matrix of the system into row-echelon form or reduced row-echelon form. This method is particularly useful for larger systems.

Steps:

1. Write the augmented matrix: Represent the system of equations as an augmented matrix.

2. Transform to row-echelon form: Use elementary row operations to transform the matrix into row-echelon form. The elementary row operations are:

   • Swapping two rows.
   • Multiplying a row by a non-zero constant.
   • Adding a multiple of one row to another.

3. Transform to reduced row-echelon form (optional): Continue using elementary row operations to transform the matrix into reduced row-echelon form.

4. Back-substitution: Read the solution from the row-echelon or reduced row-echelon form of the matrix.

Example:

Solve the following system:

• x + y + z = 6 ...(1)
• 2x - y + z = 3 ...(2)
• x + 2y - z = 2 ...(3)

Solution:

1. Write the augmented matrix:

   [ 1  1  1 | 6 ]
   [ 2 -1  1 | 3 ]
   [ 1  2 -1 | 2 ]
   

2. Transform to row-echelon form:

   • R2 -> R2 - 2R1:

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 | -9 ]
     [ 1  2 -1 | 2 ]
     

   • R3 -> R3 - R1:

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 | -9 ]
     [ 0  1 -2 | -4 ]
     

   • R2 <-> R3:

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0 -3 -1 | -9 ]
     

   • R3 -> R3 + 3R2:

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0  0 -7 | -21 ]
     

3. Transform to reduced row-echelon form:

   • R3 -> R3 / -7:

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0  0  1 | 3 ]
     

   • R2 -> R2 + 2R3:

     [ 1  1  1 | 6 ]
     [ 0  1  0 | 2 ]
     [ 0  0  1 | 3 ]
     

   • R1 -> R1 - R3:

     [ 1  1  0 | 3 ]
     [ 0  1  0 | 2 ]
     [ 0  0  1 | 3 ]
     

   • R1 -> R1 - R2:

     [ 1  0  0 | 1 ]
     [ 0  1  0 | 2 ]
     [ 0  0  1 | 3 ]
     

4. Read the solution:

   From the reduced row-echelon form, we can directly read the solution:

   x = 1, y = 2, and z = 3.

4. Matrix Methods

Matrix methods provide a powerful and efficient way to solve systems of equations, especially for larger systems. These methods involve representing the system as matrices and using matrix operations to find the solution. Two common matrix methods are using inverse matrices and Cramer's Rule.

a. Using Inverse Matrices

Steps:

1. Write the system in matrix form: Represent the system of equations as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
2. Find the inverse of matrix A: Calculate the inverse of the coefficient matrix A, denoted as A^(-1).
3. Solve for X: Multiply both sides of the equation AX = B by A^(-1) to solve for X: X = A^(-1)B.

Example:

Solve the following system:

• x + y + z = 6
• 2x - y + z = 3
• x + 2y - z = 2

Solution:

1. Write the system in matrix form:

   A =

   [ 1  1  1 ]
   [ 2 -1  1 ]
   [ 1  2 -1 ]
   

   X =

   [ x ]
   [ y ]
   [ z ]
   

   B =

   [ 6 ]
   [ 3 ]
   [ 2 ]
   

   So, AX = B is:

   [ 1  1  1 ] [ x ] = [ 6 ]
   [ 2 -1  1 ] [ y ] = [ 3 ]
   [ 1  2 -1 ] [ z ] = [ 2 ]
   

2. Find the inverse of matrix A:

   The inverse of matrix A is:

   A^(-1) =

   [ -1/3   1   -2/3 ]
   [  1/2  -1/3  1/6 ]
   [  1/6   1/3  -1/2 ]
   

3. Solve for X:

   X = A^(-1)B

   [ x ]   [ -1/3   1   -2/3 ] [ 6 ]
   [ y ] = [  1/2  -1/3  1/6 ] [ 3 ]
   [ z ]   [  1/6   1/3  -1/2 ] [ 2 ]
   

   [ x ]   [ (-1/3)*6 + 1*3 + (-2/3)*2 ]   [ -2 + 3 - 4/3 ]   [ 1 - 4/3 ]   [ -1/3*3 ]   [ 1 ]
   [ y ] = [ (1/2)*6 + (-1/3)*3 + (1/6)*2 ] = [ 3 - 1 + 1/3 ] = [ 2 + 1/3 ] = [ 7/3 ] = [ 2 ]
   [ z ]   [ (1/6)*6 + (1/3)*3 + (-1/2)*2 ]   [ 1 + 1 - 1 ]   [ 2 - 1 ]   [ 1 ]   [ 3 ]
   

   Thus, x = 1, y = 2, and z = 3.

b. Cramer's Rule

Cramer's Rule is a method that uses determinants to solve systems of linear equations.

Steps:

1. Calculate the determinant of the coefficient matrix (D): Find the determinant of matrix A.

2. Calculate the determinants Dx, Dy, and Dz: Replace the x, y, and z columns of matrix A with the constant matrix B, and calculate the determinants of these new matrices.

3. Solve for x, y, and z: Use the following formulas:

   • x = Dx / D
   • y = Dy / D
   • z = Dz / D

Example:

Solve the following system:

• x + y + z = 6
• 2x - y + z = 3
• x + 2y - z = 2

Solution:

1. Calculate the determinant of the coefficient matrix (D):

   A =

   [ 1  1  1 ]
   [ 2 -1  1 ]
   [ 1  2 -1 ]
   

   D = 1((-1)(-1) - 12) - 1(2*(-1) - 11) + 1(22 - (-1)*1) = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1) = 1(-1) - 1(-3) + 1(5) = -1 + 3 + 5 = 7

2. Calculate the determinants Dx, Dy, and Dz:

   Dx =

   [ 6  1  1 ]
   [ 3 -1  1 ]
   [ 2  2 -1 ]
   

   Dx = 6((-1)(-1) - 12) - 1(3*(-1) - 12) + 1(32 - (-1)*2) = 6(1 - 2) - 1(-3 - 2) + 1(6 + 2) = 6(-1) - 1(-5) + 1(8) = -6 + 5 + 8 = 7

   Dy =

   [ 1  6  1 ]
   [ 2  3  1 ]
   [ 1  2 -1 ]
   

   Dy = 1(3*(-1) - 12) - 6(2(-1) - 11) + 1(22 - 3*1) = 1(-3 - 2) - 6(-2 - 1) + 1(4 - 3) = 1(-5) - 6(-3) + 1(1) = -5 + 18 + 1 = 14

   Dz =

   [ 1  1  6 ]
   [ 2 -1  3 ]
   [ 1  2  2 ]
   

   Dz = 1((-1)2 - 32) - 1(22 - 31) + 6(2*2 - (-1)*1) = 1(-2 - 6) - 1(4 - 3) + 6(4 + 1) = 1(-8) - 1(1) + 6(5) = -8 - 1 + 30 = 21

3. Solve for x, y, and z:

   • x = Dx / D = 7 / 7 = 1
   • y = Dy / D = 14 / 7 = 2
   • z = Dz / D = 21 / 7 = 3

Thus, x = 1, y = 2, and z = 3.

Practical Tips and Considerations

• Check Your Work: Always substitute your solution back into the original equations to verify that it satisfies all three.
• Choose the Right Method: The elimination method is often the most straightforward for systems with simple coefficients. Substitution can be useful when one equation is already solved for a variable. Gaussian elimination and matrix methods are more efficient for larger systems.
• Be Organized: Keep your work neat and organized to avoid errors. Clearly label each step and equation.
• Watch for Special Cases: Be aware of systems with no solution or infinite solutions. If you encounter a contradiction (e.g., 0 = 1) during the solution process, the system has no solution. If you end up with fewer equations than variables, the system may have infinite solutions.

Real-World Applications

Systems of equations with three variables are used in various fields, including:

• Engineering: Solving for forces in structural analysis, designing electrical circuits.
• Economics: Modeling supply and demand, analyzing market equilibrium.
• Physics: Calculating motion, analyzing forces and energy.
• Computer Graphics: 3D modeling and rendering.
• Statistics: Regression analysis.

Conclusion

Solving systems of equations with three variables is a fundamental skill with numerous applications. By understanding the different methods available—substitution, elimination, Gaussian elimination, and matrix methods—and practicing regularly, you can become proficient in solving these systems. Remember to check your work and stay organized to avoid errors. Whether you're a student tackling algebra problems or a professional solving real-world challenges, mastering this skill will undoubtedly be valuable.